Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: Daiana on January 13, 2018, 12:34:02 PM
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The questions are attached
Ignore the word: in e-mail
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The questions are attached
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For (2) it's rather methyl-tert-butyl.
(6) is impossible. A cycle has two hydrogen atoms less than a straight or branched alkane, so it's not an isomer.
The names of dibromo look fine to me.
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Thank you a lot! I don't know how to thank you )))
How about 14.44 and that question about 2 squares.
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(6) you said that 6 is impossible. However, with CH3 it has the same number of hydrogens like the chains above. I mean that if we took the complete formula - we would get 12 hydrogens.
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You can always use http://www.chemspider.com/FullSearch.aspx (http://www.chemspider.com/FullSearch.aspx) for checking IUPAC name.
all you need to do is just draw the structure.
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(6) has 2*4+1+3=12 hydrogen atoms. As opposed, a straight or ramified ether with 6 carbons has 6*2+2=14 hydrogens.
You don't need to count every hydrogen atom, which is prone to errors. It suffices to know that straight or ramified alkanes have 2*C+2 hydrogens, and that every ring or double bond takes 2*H fewer (triple bonds 4*H, lone benzene rings 8*H etc).
That is, a ring can't be an isomer of a straight alkane. It's an isomer of an alkene.
Imagine that you close a ring from a straight alkane by binding both end C. Each carries 3*H so no bond is possible. Only by removing one H on each C can you make the new C-C bond.
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The use of two squares for a cyclobutyl group may be a misprint of some kind. I don't recall ever seeing that depiction before.